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Thread: [SOLVED] Function Question

  1. #1
    Member looi76's Avatar
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    [SOLVED] Function Question

    Question:
    The function $\displaystyle f : x \mapsto x^2 + 5x + 8$ is defined for the domain $\displaystyle x \leq a$, where a is a constant.
    (i) Express $\displaystyle x^2 + 5x + 8$ in the form $\displaystyle (x + p)^2 + q$.
    (ii) Find the smallest value of a for which $\displaystyle f$ has an inverse.
    (iii) Find the domain of $\displaystyle f^{-1}$ corresponding to this value of $\displaystyle a$.


    Attempt:
    (i) $\displaystyle f(x) = x^2 + 5x + 8$

    $\displaystyle = (x + \frac{5}{2})^2 - \left(\frac{5}{2}\right)^2 - 8$

    $\displaystyle = (x + \frac{5}{2})^2 + \frac{7}{4}$

    $\displaystyle p = \frac{5}{2} \mbox{ and }q = \frac{7}{4}$

    Is the domain $\displaystyle = -\frac{5}{2}$ and range $\displaystyle = \frac{7}{4}$ ?

    $\displaystyle = (x + \underbrace{\frac{5}{2}}_{Domain})^2 + \underbrace{\frac{7}{4}}_{Range}$
    Last edited by looi76; May 4th 2008 at 12:26 PM.
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  2. #2
    Kai
    Kai is offline
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    Draw the graph, easier to know domain and range,

    And remember, domain of f(x)= range of finverse(x)
    Range of f(x) = Domain of finverse(x)
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  3. #3
    MHF Contributor
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    Quote Originally Posted by looi76 View Post
    Question:
    The function $\displaystyle f : x \mapsto x^2 + 5x + 8$ is defined for the domain $\displaystyle x \leq a$, where a is a constant.
    (i) Express $\displaystyle x^2 + 5x + 8$ in the form $\displaystyle (x + p)^2 + q$.
    (ii) Find the smallest value of a for which $\displaystyle f$ has an inverse.
    (iii) Find the domain of $\displaystyle f^{-1}$ corresponding to this value of $\displaystyle a$.

    Attempt:
    (i) $\displaystyle f(x) = x^2 + 5x + 8$

    $\displaystyle = (x + \frac{5}{2})^2 - \left(\frac{5}{2}\right)^2 - 8$

    $\displaystyle = (x + \frac{5}{2})^2 + \frac{7}{4}$

    $\displaystyle p = \frac{5}{2} \mbox{ and }q = {7}{4}$

    Is the domain $\displaystyle = -\frac{5}{2}$ and range $\displaystyle = \frac{7}{4}$ ?

    $\displaystyle = (x + \underbrace{\frac{5}{2}}_{Domain})^2 + \underbrace{\frac{7}{4}}_{Range}$
    For the function $\displaystyle (x + \frac{5}{2})^2 + \frac{7}{4}$ the domain is $\displaystyle x \leq -\frac{5}{2}$ and the range is $\displaystyle y \geq \frac{7}{4}$.
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